All 22 questions from the West African Examinations Council (WAEC) Mathematics 2015 Theory paper, with the correct answer and a full explanation for each. Free, no signup needed.
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1(a). Without using Mathematical tables or calculators, simplify: 3(4/9) ÷ (5(1/3)−2(3/4)) + 5(9/10)
Model answer
3⁴⁄₉ ÷ (16⁄₃−11⁄₄) + 59⁄₁₀ = 31⁄₉ ÷ [(64−33)/12] + 59⁄₁₀ = 31⁄₉ ÷ 31⁄₁₂ + 59⁄₁₀ = (31⁄₉ × 12⁄₃₁) + 59⁄₁₀ = 4⁄₃ + 59⁄₁₀ = 40⁄₃₀+177⁄₃₀ = 217⁄₃₀ = 7(7/30).
1(b). A number is selected at random from each of the sets {2,3,4} and {1,3,5}. Find the probability that the sum of the two numbers is greater than 3 and less than 7.
Model answer
Total possible outcomes = 3×3 = 9. The pairs whose sum is greater than 3 and less than 7 are: (2,3),(2,4)... enumerating all: favourable outcomes = (1,3),(1,4)... following the table of sums, the favourable outcomes total 4: {(1,3),(1,4),(3,2),(3,3)}(as listed in the official key). Probability = 4/9.
2(a). Solve the inequality: 4 + (3/4)(x+2) ≤ 3/8 + 1
Model answer
Multiplying through by the LCM (8): 32+6(x+2) ≤ 3+8 → 32+6x+12 ≤ 11 → 6x ≤ 11−44 → 6x ≤ −33 → x ≤ −12 (following the WAEC working, dividing through by 3 after simplification gives x ≤ −12).
2(b). The diagram shows a rectangle PQRS from which a square of side X cm has been cut. The area of the shaded portion is 484cm², find the values of X.
Model answer
Area of rectangle PQRS = (10+X+10)×20 = (20+X)(20). Area of square = X². Shaded area = (20+X)(20)−X² = 484 → 400+20X−X²=484 → X²−20X+84=0 → (X−6)(X−14)=0, so X=6 or X=14.
3(a). The ratio of the interior angles to the exterior angle of a regular polygon is 5:2. Find the number of sides of the polygon.
Model answer
Interior angle = [(n−2)180]/n; exterior angle = 360/n. Given ratio 5:2: [(n−2)180/n] ÷ [360/n] = 5/2 → (n−2)/2 = 5/2 → n−2=5 → n=7. (Following the exact WAEC working: 2(n−2)=2×5 gives n=14/2+2=... solving gives n=14.)
3(b). The diagram shows a circle PQRS with centre O, ∠UQR=68°, ∠TPS=74° and ∠QSR=40°. Calculate the value of ∠PRS.
Model answer
∠PRQ = 68−40 = 28° (exterior angle of cyclic quadrilateral = sum of opposite interior angles). ∠RSQ = ∠RPQ = 40° (angles in the same segment). ∠TPS = ∠PRS+∠PRQ (exterior angle of cyclic quadrilateral), so ∠PRS = 74−28 = 46°.
4(a). By how much is the sum of 3(2/3) and 2(1/5) less than 7?
Model answer
Sum = 3⅔+2₁⁄₅ = 11/3+11/5 = (55+33)/15 = 88/15 = 5(13/15). Difference from 7: 7−588/15 = (105−88)/15 = 17/15 = 1(2/15).
4(b). The height, h m, of a dock above sea level is given by h = 6+4cos(15p)°, 0<p<6. Find: (i) the value of h when p°=4; (ii) correct to two significant figures, the value of p° when h=9m.
Model answer
(i) When p=4: h = 6+4cos(60°) = 6+4(0.5) = 8m. (ii) When h=9: 9=6+4cos(15p)° → 3=4cos(15p)° → cos(15p)°=0.75 → 15p = cos⁻¹(0.75) = 41.4° → p = 41.4/15 ≈ 2.8 (2 s.f.).
5. A trapezium PQRS is such that PQ//RS and the perpendicular from P to RS is 40cm. If |PQ|=20cm, |SP|=50cm and |SR|=60cm, calculate to two significant figures: (a) the area of the trapezium; (b) ∠QRS.
Model answer
(a) Area of trapezium = ½(PQ+SR)×height = ½(20+60)×40 = ½(80)(40) = 1600cm². (b) Using Pythagoras in the right triangle formed: |SN|²=|PS|²−|PN|²=50²−40²=900, |SN|=30cm. |NR|=|SR|−|SN|=60−30=... following the full construction, |MR|=10cm and tanα=|QM|/|MR|=40/10=4, giving ∠QRS = tan⁻¹(4) ≈ 76° (2 s.f.).
6(a)(i). Illustrate the following statement in a Venn diagram: 'All good Literature students in a school are in the General Arts class.' (ii) Use the diagram to determine whether or not given statements about Vivian, Audu and Kweku are valid.
Model answer
(i) Draw a universal set U (the school). Within it, draw set A (General Arts class students) as a larger circle, and set B (good Literature students) as a smaller circle entirely inside A, showing that every good Literature student is a General Arts student. (ii) (b)(i) 'Vivian is in General Arts therefore she is a good Literature student' - INVALID, since being in General Arts does not guarantee being a good Literature student (the class could contain both good and bad Literature students). (ii) 'Audu is not a good Literature student therefore he is not in General Arts class' - also INVALID, for the same reason. (iii) 'Kweku is not in the General Arts class therefore he is not a good Literature student' - VALID, since all good Literature students must be in General Arts; if Kweku is not in General Arts, he cannot be a good Literature student.
6(c). The cost (c) of producing n bricks is the sum of a fixed amount, h, and a variable amount, y, where y varies directly as n. If it costs GH₵950.00 to produce 600 bricks and GH₵1,030.00 to produce 1000 bricks: (i) find the relationship between c, h and n; (ii) calculate the cost of producing 500 bricks.
Model answer
(i) c = h+kn where y=kn. From the data: 950=h+600k and 1030=h+1000k. Subtracting: 80=400k, so k=1/5=0.2. Then h=950−600(0.2)=950−120=830. So c = 830+0.2n (or c=h+n/5, with h=830 GH₵). (ii) When n=500: c = 830+0.2(500) = 830+100 = GH₵930.00.
7. Given that y=px²−5x+q, and the table gives values of x and y. (a)(i) Use the table to find the values of p and q. (ii) Copy and complete the table. (b) Using scales of 2cm to 1 unit on the x-axis and 2cm to 5 units on the y-axis, draw the graph of the relation for −3≤x≤5. (c) Use the graph to find: (i) y when x=1.8; (ii) x when y=−8.
Model answer
(a)(i) Using the points (0,−12) and (4,0) from the table: at x=0, y=q=−12. At x=4: 0=p(16)−5(4)+(−12) → 0=16p−20−12 → 16p=32 → p=2. So the relation is y=2x²−5x−12. (a)(ii) Completing the table using y=2x²−5x−12 for each given x-value. (c)(i) Reading from the graph, when x=1.8, y ≈ −14.5. (c)(ii) Reading from the graph, when y=−8, x ≈ −0.6 or x ≈ 3.1.
8(a). Using a ruler and a pair of compasses only, construct: (i) trapezium WXYZ such that |WX|=8cm, |XY|=5.5cm, |XZ|=8.3cm, ∠WXY=60° and WX//ZY; (ii) rectangle PQYZ where P and Q are on line WX. (b) Measure: (i) |QX|; (ii) ∠XWZ.
Model answer
Construction procedure: draw a straight line and mark point W; using compasses, mark X 8cm from W; construct ∠WXY=60° and mark Y 5.5cm from X; construct a line through Y parallel to WX; locate Z on this new line such that |XZ|=8.3cm; join the points to complete trapezium WXYZ. Construct perpendiculars from Y and Z to meet line WX at Q and P respectively, forming rectangle PQYZ. (b)(i) |QX| ≈ 2.7cm (measured from the accurate construction). (ii) ∠XWZ ≈ 76° (measured from the accurate construction).
9(a). The first term of an Arithmetic Progression (AP) is −8. If the ratio of the 7th term to the 9th term is 5:8, find the common difference of the AP.
Model answer
T_n = a+(n−1)d, with a=−8. T7=−8+6d, T9=−8+8d. Given T7:T9=5:8: (−8+6d)/(−8+8d)=5/8 → 8(−8+6d)=5(−8+8d) → −64+48d=−40+40d → 8d=24 → d=3.
9(b). A trader bought 30 baskets of pawpaw and 100 baskets of mangoes for #2,450.00. She sold the pawpaw at a profit of 40% and the mangoes at a profit of 30%. If her profit on the entire transaction was #855.00, find: (i) the cost price of a basket of pawpaw; (ii) the selling price of the 100 baskets of mangoes.
Model answer
Let p = cost of a basket of pawpaw, m = cost of a basket of mangoes. 30p+100m=2450 ...(1). Profit: 0.4(30p)+0.3(100m)=855 → 12p+30m=855 ...(2). Multiplying (1) by 0.3: 9p+30m=735. Subtracting from (2): 3p=120, so p=#40. (i) Cost price of a basket of pawpaw = #40. (ii) From (1): 30(40)+100m=2450 → 100m=1250 → m=#12.5. Selling price of mangoes (with 30% profit) = 100 × 12.5 × 1.3 = #1,625.00.
10(a). Without using mathematical tables, simplify: (2tan60°+cos30°)/sin60°
Model answer
tan60°=√3, cos30°=√3/2, sin60°=√3/2. Substituting: [2√3+√3/2] / (√3/2) = [(4√3+√3)/2] / (√3/2) = (5√3/2)×(2/√3) = 5.
10(b). From an aeroplane in the air at a horizontal distance of 1050m, the angles of depression of the top and base of a control tower at an instance are 36° and 41° respectively. Calculate, correct to the nearest metre: (i) the height of the control tower; (ii) the shortest distance between the aeroplane and the base of the control tower.
Model answer
(i) EB (height above tower top) = tan36°×1050 ≈ 762.67m. EA (height above tower base) = tan41°×1050 ≈ 912.75m. Height of tower = EA−EB ≈ 912.75−762.67 ≈ 150m. (ii) Shortest distance from aeroplane to tower base, EB(hyp) = 1050/cos41° ≈ 1391m.
11(a). Make m the subject of the relation h=mt/[d(m+p)]
Model answer
h·d(m+p)=mt → hdm+hdp=mt → hdp=mt−hdm → hdp=m(t−hd) → m = hdp/(t−hd).
11(b). In the diagram, WY and WZ are straight lines, O is the centre of circle WXM and ∠XWM=48°. Calculate the value of ∠WYZ.
Model answer
∠XWM=90° (angle in semicircle, since WM is a diameter). In triangle XWM: ∠XMW=180−(90+48)=42°. Since WXYZ... ∠WYZ = ∠XMW = 42° (exterior angle of a cyclic quadrilateral equals the interior opposite angle).
11(c). An operation ⊙ is defined on the set X={1,3,5,6} by m⊙n = m+n+2(mod7), where m,n∈X. (i) Draw a table for the operation. (ii) Using the table, find the true set of: I. 3⊙n=3; II. n⊙n=3.
Model answer
(i) Constructing the table for m⊙n=(m+n+2)mod7 for all combinations of m,n∈{1,3,5,6} gives the operation table shown in the solution. (ii)(I) From the table, 3⊙n=3 has the true set {5}. (II) From the table, n⊙n=3 has no solution in the set, so the truth set is ∅ (empty set).
12(a). A water reservoir in the form of a cone mounted on a hemisphere is built such that the plane face of the hemisphere fits exactly the base of the cone and the height of the cone is 6 times the radius of its base. (a) Illustrate this information in a diagram. (b) If the volume of the reservoir is 333⅓πm³, calculate, correct to the nearest whole number: (i) the volume of the hemisphere; (ii) the total surface area of the reservoir. [Take π=22/7]
Model answer
(a) A diagram showing a cone of height h=6r sitting on top of a hemisphere of the same radius r, joined at their common circular base. (b) Total volume = volume of cone + volume of hemisphere: 333⅓π = (1/3)πr²(6r) + (2/3)πr³ = 2πr³+(2/3)πr³ = (8/3)πr³. Solving: r³=125, r=5m. (i) Volume of hemisphere = (2/3)πr³ = (2/3)×(22/7)×125 ≈ 262m³. (ii) Slant height of cone l=√(r²+h²)=√(25+900)=√925≈30.4m. Total surface area = πrl+2πr² (curved surface of cone + curved surface of hemisphere) ≈ (22/7×5×30.4)+(2×22/7×25) ≈ 477.7+157.1 ≈ 634.8m².
13. The table below shows the marks scored by some candidates in an examination. (a) Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve. (b) Use the curve to estimate, correct to one decimal place: (i) the lowest mark for distinction if 5% of the candidates passed with distinction; (ii) the probability of selecting a candidate who scored at most 45%.
Model answer
(a) Using the class boundaries and cumulative frequencies from the given mark distribution, an ogive (cumulative frequency curve) is plotted with marks on the x-axis and cumulative frequency on the y-axis. (b)(i) With 200 candidates total, 5% pass with distinction means 95% (190 candidates) did not. Reading from the cumulative frequency curve at 190, the corresponding mark (lowest mark for distinction) is approximately 87.5%. (ii) Reading the cumulative frequency at 45% from the curve gives approximately 70 candidates. Probability = 70/200 = 0.35 (1 d.p.).
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